Bounded number of multiple roots in a family of polynomials Let $F$ be a field and let $f(x)$ be a fixed nonconstant polynomial. Look at the family of polynomials $x^n - f(x)$, where $n$ varies. It is reasonable to assume that these have $n - C$ distinct roots (ignoring multiplicity!) at least, where $C$ is a constant depending only on $f(x)$. Any idea how to prove this?
To see that this question is nontrivial (at least for me, unless I miss something obvious) note that in characteristic $p$, over say an algebraic closure of $\mathbb{F}_p(t)$, each of the polynomials $x^{p^n} - t$ has a single root.
 A: Here is a beginning, which gives the possibilities for multiple roots in terms of $f(x)$:
Let $g_n(x)=x^n-f(x)$ so that $g_n'(x)=nx^{n-1}-f'(x)$ and $g_n'(x)-ng(x)=nf(x)-f'(x)$
Since a multiple root will also be a root of the derivative, the roots of $nf(x)-f'(x)=0$ are the only possibilities for multiple roots.
Note multiple roots do obviously occur when $f(x)=x^rh(x)$.
A: I suppose that $k$ has characteristic $0$, and that $f\not = 0$, of degree $d$, and I take the zeros in an algebraic closure of $k$. A multiple root of $x^n-f(x)$ of order $k$ is such that $x^n=f(x)$ and $n! f(x)-(n-j)!f^{(j)}(x)=0$ for $1\leq j\leq k-1$. If we suppose that $k\geq d+2$, we get for $j=d+1$ that $n! f(x)=0$, hence by $x^n=f(x)$ that $x=0$. But the order of $0$ as a root of $x^n-f(x)$ is at most $d$ for $n\geq d+1$. Hence all repeated zeros of $x^n-f(x)$ have for $n\geq d+1$ an order at most $d+1$, and they are also zeros of $P_n(x)=nf(x)-xf^{\prime}(x)$; for $n\geq d+1$, $P_n$ is a polynomial of degree $d$, thus there is at most $d$ such zeros. Hence the repeated zeros contribute (counting multiplicity) for at most $d(d+1)$ of the number of roots of $x^n-f(x)$. Thus there is at least $n-d(d+1)$ simple zeros of $x^n-f(x)$ for $n\geq d+1$. 
Edit: For $k$ of characteristic $p$, take as an example $f(x)=x^p$. For $n=mp$, we have $x^n-f(x)=x^{mp}-x^p=(x^m-x)^p$, and there exists no simple zeros.
